SOLUTION: Airport B is 300 mi from airport A at a bearing N 50°E (see the figure). A pilot wishing to fly from A to B mistakenly flies due east at 175 mi/h for 35 minutes, when he notices hi
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Question 508239: Airport B is 300 mi from airport A at a bearing N 50°E (see the figure). A pilot wishing to fly from A to B mistakenly flies due east at 175 mi/h for 35 minutes, when he notices his error
How far is the pilot from his destination at the time he notices the error? Give your answer correct to the nearest mile?
What bearing should he head his plane in order to arrive at airport B? Give your answer correct to the nearest degree?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Airport B is 300 mi from airport A at a bearing N 50°E (see the figure). A pilot wishing to fly from A to B mistakenly flies due east at 175 mi/h for 35 minutes, when he notices his error
How far is the pilot from his destination at the time he notices the error? Give your answer correct to the nearest mile?
What bearing should he head his plane in order to arrive at airport B? Give your answer correct to the nearest degree?
**
Draw a line from point A to point B. This line is 300 mi long and makes an angle of 40º from the horizontal (based on N50ºE bearing). From point A draw a horizontal line to point C. This line=(35/60)*175=102.1 mi. Draw a line connecting point C to point B. You now have a triangle with sides AB=300 mi and AC=102.1 mi and their included angle of 40º.
Using Law of cosines to find CB:
(CB)^2=(AB)^2+(AC)^2-2*AB*AC*cos 40º
(CB)^2=(300)^2+(102.1)^2-2*300*102.1*cos 40º
(CB)^2=90000+10424-46928=53497
CB=√53497=231 mi
..
Solving for bearing
CB/sin 40º=300/sin(ACB)
231/sin40º=300/sin(ACB)
sin(ACB)=(300/231)sin40º
sin(ACB)=.8348
ACB=123.4º
new heading=90-123.4=33.4º
ans:
The pilot is 231 mi from his destination at the time he notices the error.
In order to arrive at airport B, the new heading should be N33.4ºW
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